What is the integral from 2 to 3 of (20)/(sqrt(3-x)) ?

The integral from 2 to 3 of (20)/(sqrt(3-x)) is 40

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integral from 2 to 3 of (20)/(sqrt(3-x))

\int_{2}^{3} \frac{20}{3 - x} d x

40

Solution steps

\int_{2}^{3} \frac{20}{3 - x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{u} d u

= 20 \cdot \int_{1}^{0} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 20 (- \int_{0}^{1} - \frac{1}{u} d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{1}

= 20 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

20 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 20 [2 u]_{0}^{1}

= 20 [2 u]_{0}^{1}

Compute the boundaries:

2

= 20 \cdot 2

Simplify

= 40

\frac{\partial}{\partial x} (3 cos (x y))

\frac{1}{( 1 - 3 \cdot x ^{3} ) ^{3}}

f (x) = 10 x, (10, 10)

\int \frac{x - 1}{x ^{2} - 4 x + 5} d x

\frac{d y}{d x} + y = - 1, y (0) = 1

What is the integral from 2 to 3 of (20)/(sqrt(3-x)) ?
The integral from 2 to 3 of (20)/(sqrt(3-x)) is 40